1678. Goal Parser Interpretation - Complete C++ Solution
Featured Summary: The Goal Parser Interpretation problem (LeetCode 1678) requires us to parse a given command string and replace specific substrings based on a strict set of rules. By sequentially iterating through the string, we map the 'G' character to 'G', the '()' sequence to 'o', and the '(al)' sequence to 'al'. Using C++, this problem can be solved highly efficiently in O(N) time complexity by checking the current character and its subsequent characters to construct the final concatenated parsed string.
Understanding the Goal Parser Problem
In the realm of computer science, building parsers to interpret structured commands is a highly sought-after skill. You own a Goal Parser that can interpret a string command. The command consists of an alphabet of "G", "()" and/or "(al)" in some order. The Goal Parser will explicitly interpret "G" as the string "G", "()" as the string "o", and "(al)" as the string "al". The interpreted strings are then concatenated in the original order to form the final result.
Given the string command, return the Goal Parser's interpretation of command.
This problem tests your fundamental ability to process text linearly, identify patterns, and cleanly generate a modified string output without encountering out-of-bounds errors or unnecessary computational overhead.
Example Scenarios for the Parser
Let's look at several examples to fully understand the input and output requirements:
Example 1:
Input: command = "G()(al)"
Output: "Goal"
Explanation: The Goal Parser interprets the command sequentially as follows:
G -> G
() -> o
(al) -> al
The final concatenated result is strictly "Goal".Example 2:
Input: command = "G()()()()(al)"
Output: "Gooooal"
Explanation: Multiple "()" sequences are successfully transformed into "o" characters.Example 3:
Input: command = "(al)G(al)()()G"
Output: "alGalooG"
Explanation: The parser seamlessly processes mixed sequences and single characters.Optimal C++ Solution for Goal Parser Interpretation
Below is the highly optimized C++ implementation to solve this string manipulation challenge. We utilize a straightforward linear scan algorithm that builds the output string efficiently on the fly.
class Solution {
public:
string interpret(string command) {
string result = "";
for(int i = 0; i < command.size(); i++){
if(command[i] == 'G') {
result += "G";
}
else if(command[i] == '(' && command[i+1] == ')') {
result += "o";
i++;
}
else if(command[i] == '(' && command.substr(i,4) == "(al)"){
result += "al";
i += 3;
}
}
return result;
}
};Detailed Algorithm Explanation and Complexity Analysis
The C++ implementation above focuses on creating a new result string and appending the correctly interpreted characters as we traverse the original command string from left to right. Let us thoroughly break down the programming logic:
- Iterative Linear Scan: We loop through each character in the string using a standard
forloop. The indexikeeps track of our current position. - Handling 'G': If the current character is 'G', we simply append 'G' to our continuously growing result string.
- Handling '()': If we encounter an opening parenthesis '(', we immediately check the next character. If it is a closing parenthesis ')', we definitively know we have found the '()' sequence. We append the character 'o' to the result and increment our loop index
iby 1 to effectively skip the closing parenthesis. - Handling '(al)': If the sequence starting with an opening parenthesis is not '()', we utilize the
substrfunction to check if the substring of length 4 starting at the current index matches '(al)'. If it does, we append 'al' to the result and increment the loop indexiby 3 to correctly skip the remaining characters 'a', 'l', and ')'.
This linear scanning approach ensures that we only visit each character a constant number of times. Consequently, this results in an optimal O(N) time complexity, where N is the length of the string, and O(N) space complexity required exclusively for generating the newly created string. There are no nested loops or complex data structures needed.
Frequently Asked Questions (FAQ)
What is the exact time complexity for the Goal Parser Interpretation solution?
The time complexity is firmly O(N), where N is the total length of the command string. We iterate through the string exactly once. Furthermore, the substr check for a fixed length of 4 takes constant time O(1). This makes the algorithm highly efficient and capable of scaling seamlessly to maximum input constraints.
Can we solve Goal Parser Interpretation using standard regular expressions?
Yes, regular expressions (Regex) can easily be used to replace '()' with 'o' and '(al)' with 'al' globally across the string. However, the manual string iteration approach in C++ demonstrated above is generally much faster. It explicitly avoids the significant computational overhead and memory footprint associated with invoking a robust regex engine.
Why do we need to increment the loop index by 3 for the (al) sequence?
When we successfully encounter the explicit sequence '(al)', it undeniably occupies 4 characters in the original string. Since our overarching for loop automatically increments the index by 1 at the end of each iteration, we must manually add an additional 3 to skip the remaining characters of the sequence. This effectively advances the parser past the entire sequence in one swift movement.
Conclusion and Next Steps
Understanding how to efficiently parse and dynamically interpret command strings is an absolutely fundamental skill in software engineering. The Goal Parser Interpretation problem provides exceptional, practical experience for mastering string traversal, index manipulation, and conditional logic within C++. By actively employing a straightforward linear scan, we guarantee optimal performance that avoids unnecessary iterations. Keep practicing similar algorithmic sequence manipulation problems on platforms like LeetCode to further refine your technical coding abilities and excel in software engineering interviews!